1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pushing a cabinet on a rough floor

  1. Jun 19, 2013 #1
    1. The problem statement, all variables and given/known data
    You are pushing a filing cabinet horizontally across a rough floor (μk = 0.37) in a straight line at a constant speed. Which of the following statements about the magnitudes of the forces acting on the filing cabinet are correct?

    1) The force that you exert on the filing cabinet will be more than its weight?

    2) The force that you exert on the filing cabinet will be equal to the frictional force on the cabinet

    3) The force that you exert on the filing cabinet will be equal to its weight

    4) The force that you exert on the filing cabinet will be less than its weight

    5) The force that you exert on the filing cabinet will be more than the frictional force on the cabinet

    6) If you exerted twice the force, the cabinet would slide at a constant speed that is twice the original value

    7) If you exerted twice the force, the cabinet would accelerate across the floor




    2. Relevant equations



    3. The attempt at a solution

    1 I know is false, because cabinet is moving at a constant speed, hence [itex]F = μMg[/itex]

    2 is true because [itex]F = μMg[/itex] (Constant speed)

    3 is false. Less than it's weight. Because μ is less than 1

    4 true. Because μ is less than 1

    5 I have a doubt about this one. Will it be true? Because, to keep it at constant speed A force constant force greater than frictional force is required

    6 Not sure

    7 Not sure.

    Please correct me where I'm wrong and give me an idea about 6th and 7th
     
  2. jcsd
  3. Jun 19, 2013 #2

    Curious3141

    User Avatar
    Homework Helper

    1. Correct, because ##\mu_k < 1##. 2 to 4 are also correct.

    5. How can this be true if 2 is also true?

    6. Would there be an unbalanced net force on the cabinet in this case? What does F = ma tell you?

    7. See 6.
     
  4. Jun 19, 2013 #3
    I'm assuming 6 is true, 7 is false.
    Because if it's moving at twice the constant speed, it's obviously not accelerating.
    I had the exact same doubt about 5th. But if it wasn't greater than frictional force, and just equal to it, wouldn't it just stay in equilibrium?
     
  5. Jun 19, 2013 #4

    Curious3141

    User Avatar
    Homework Helper

    No! Why would you assume that?

    Perhaps it's better to start at the start. What do Newton's first two laws of motion state?

    So what exactly is equilibrium? Does it necessarily mean a body is at rest? Can a body be in motion and still remain in equilibrium?
     
  6. Jun 19, 2013 #5
    I'm a terrible teacher
     
  7. Jun 19, 2013 #6
    No, it doesn't mean body is at rest, here equilibrium implies to keep the body at a constant speed AGAINST the frictional force. So does that mean a force that would be equal to frictional force be able to do that?

    Also, I'm still not clear about 6th and 7th. If twice the force exerted is applied constantly, it'd slide at a constant speed. (according to first law). But it's no where mentioned that it is applied constantly. Does that mean if twice the force was exerted, the applied force would decrease? Hence, making 6th false and 7th true (because it is decelerating)
     
  8. Jun 19, 2013 #7

    Curious3141

    User Avatar
    Homework Helper

    Constant velocity (which also implies constant speed), so yes, correct. If the force you exert is equal in magnitude and exactly opposite in direction to the frictional force, the body will be in equilibrium. In this case, equilibrium means the cabinet moves in a straight line at constant speed.

    If twice the force were exerted, what would be the *net* force on the object?

    The frictional force remains constant at ##\mu_kMg##.

    Initially, the force you applied was the same: ##\mu_kMg##. The net force was zero. This meant the cabinet *was* in equilibrium, travelling at constant velocity.

    Now, the force you're applying is ##2\mu_kMg##. So what's the net force on the cabinet now?

    What does ##F = ma## tell you?

    When ##a## is nonzero, can an object move with constant velocity?
     
  9. Jun 19, 2013 #8
    Are you implying

    [tex]F - μkMg= Ma [/tex]?

    Hence, [tex] 2μkMg - μkMg = Ma [/tex]

    Therefore, a = μkg?
     
  10. Jun 19, 2013 #9
    Okay, so are 6 and 7 true? I just worked out again, and I get them as true.
     
  11. Jun 19, 2013 #10
    But again, 6 says " the cabinet would slide at a constant speed that is twice the original value " . Do they mean twice the original value of speed? How do I prove that? I'm clear about 7th.
     
  12. Jun 19, 2013 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you think the 7th is true, then the 6th certainly can't be true. What do you think?
     
  13. Jun 19, 2013 #12
    Yup, you're right. I still can't understand the 6th one though. Rest is very clear.
     
  14. Jun 20, 2013 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If 6th isn't true then you don't have to understand it. Do you? The 7th says the cabinet will accelerate, so it can't have a constant velocity.
     
  15. Jun 20, 2013 #14

    Curious3141

    User Avatar
    Homework Helper

    Yes. ##a = \mu_kg##. (BTW, you should learn to do subscripts properly in LaTex, otherwise it looks like k is a constant you're multiplying everything by. Here, the subscript k refers to "kinetic friction").

    So if there's a nonzero acceleration, 7 is clearly true. Agreed?

    Can 6 be true if 7 is true?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Pushing a cabinet on a rough floor
Loading...